The lifespans of lions in a particular zoo are normally distributed. The average lion lives $13.1$ years; the standard deviation is $1.9$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living less than $18.8$ years.
Solution: $13.1$ $11.2$ $15$ $9.3$ $16.9$ $7.4$ $18.8$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $13.1$ years. We know the standard deviation is $1.9$ years, so one standard deviation below the mean is $11.2$ years and one standard deviation above the mean is $15$ years. Two standard deviations below the mean is $9.3$ years and two standard deviations above the mean is $16.9$ years. Three standard deviations below the mean is $7.4$ years and three standard deviations above the mean is $18.8$ years. We are interested in the probability of a lion living less than $18.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the lions will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the lions will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $7.4$ years and the other half $({0.15\%})$ will live longer than $18.8$ years. The probability of a particular lion living less than $18.8$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.